GATE CSE 2010 | Question: 47

Question

Suppose computers $A$ and $B$ have $IP$ addresses $10.105.1.113$ and $10.105.1.91$ respectively and they both use same netmask $N$. Which of the values of $N$ given below should not be used if $A$ and $B$ should belong to the same network?

• A. $255.255.255.0$
• B. $255.255.255.128$
• C. $255.255.255.192$
• D. $255.255.255.224$

Comments

$A = \textbf{10.105.1.(01}110001)_2$
$B = \textbf{10.105.1.(01}011011)_2$

The first 26 bits of $A$ and $B$ match, therefore netmask $N$ can range from $/0$ to $/26$ for $A$ and $B$ to be in the same network.
Option $D.$ is $/27$ which is outside the range of $N$
$\therefore \  D.$ is the answer.

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To solve this simply we do AND operation for given 2 ip add with the network mask N in option and check if the network id we got is same or not .

Answer is D

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Silly mistake done that i haven't clearly understood that which subnet is not useful here

Answer 1

D is correct answer because:

When we perform $AND$ operation between $IP$ address $10.105.1.113$ and $255.255.255.224$ result is  $10.105.1.96$

When we perform $AND$ operation between $IP$ address  $10.105.1.91$ and $255.255.255.224$ result is $10.105.1.64$

$10.105.1.96$ and $10.105.1.64$ are different network so $D$ is correct answer.

Comments

The question says which of the values of N given below should not be used if A and B should belong to the same network. According to option 4 A and B do not belong to same network hence the option should not be considered only. If i choose answer as A where least numbers of bits are used to represent subnet mask. While in option B and C more nos of bits are used in Subnet mask. Hence answer should be A why D is selected as an option?

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The question says which of the values of N given below should not be used if A and B should belong to the same network.

As you wrote yourself. the question asks which value of netmask N should not be used. 

We are to find which netmask among the given options make it so that A and B are not in the same network (from each others' perspective).

 

IP	Last Byte
$\text{IP}_A$	0111 0001
$\text{IP}_B$	0101 1011

 

As we can see, the first two most-significant-bits of the last byte of the IP addresses of A, B are matching.

Any netmask upto that length will result in A and B being in the same network.

Any netmask greater than that length will result in A and B not being in the same network (what the question is actually asking from us).

Out of all the options, option D is the only netmask which will do so. That's why it's the answer.

Answer 2

ANSWER : OPTION D
 

Method-  

128  64  32  16  8  4  2  1

  0     1   1    1   0  0  0  1  --- (113 last octet of first address)
  
  0     1    0   1   1  0   1  1  ----(91 last octet of second address)

Now, first MSB and second bit matching but 3rd bit is different.

so , we need 3  continos one's in mask, for different network.
So mask will be 255.255.255.224
 

@Arjun Sir please check , this method is also right or not?

Comments

Good method

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I apply the same...right method...

Answer 3

The last octets of IP addresses of A and B 

A:113 (01110001) and

B: 91 (01011011). 

The netmask in option (D) has first three bits are 111  in last octet. If netmask has first 3 bits set, then these bits must be same in A and B, but that is not the case. we can say option (D) is not a valid netmask because doing binary ‘&’ of it with addresses of A and B doesn’t give the same network address. It must be same address as A and B are on same network. 

Comments

good explanation.

Answer 4

To solve these kind of questions, first approach should be finding network ID

Now given all the subnet masks, try to calculate Network ID for both the IP addresses and you will find out that

last octet of both the IP addresses are 113 and 91 respectively

With 224 in the last octet of Subnet mask, we are not able to get same Network ID as

113 is 01101111 and

  91 is 01011011 

with 0 mask we get 0 as network id

with 128 we get 0 as network id

with 192 we get 64 as network id

but with 224 we get different network ids

Hence option with 224 in the last octet is the correct answer.

 

 

Answer 5

Ans 

Answer 6

The obvious method might be to AND the given IP's with the mask and check if they give the same result or not, this is good but not the fastest way to go around...

First we can make some observations regarding the given two IP's: 10.105.1.113 and 10.105.1.91

Clearly the first three octets are same so any netmask $\leq{}/24$ will consider them on the same network. Now we only need to check the masks:

A. 255.255.255.0 = \24

B. 255.255.255.128 = \25

C. 255.255.255.192 = \26

D. 255.255.255.224 = \27

1 is clearly not the correct option, for the rest of the options we need only check at maximum the first 3 bits of the last octet to confirm if they are part of the network:
113 = 0111 0001

91  =  0101 1011
Clearly upto the 2nd bit they match so we any netmask greater than 24+2 = 26 will not consider them in the same network, making D the answer. 
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